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  • Vemuri's Brain Teaser-3

    mathbyvemuri

    Member

    Updated: Oct 25, 2024
    Views: 1.1K
    On 1st May, Mr. X arrived in a new city and was looking for a place to stay. He met a landlard who offered to rent his apartment at a reasonable price but wanted Mr. X to pay the rent on a daily basis. Mr. X had a silver bar of 31 inches and an inch of the silver bar was exactly equal to a day’s rent. He agreed to pay an inch of the silver bar towards the daily rent. Mr. X wanted to make minimum number of pieces of silver bar but did not want to pay any advance rent. How many pieces did he make?
    (A)5 (B)8 (C)16 (D)20 (E)31
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  • Dancer_Engineer

    Member • May 16, 2012

    I tried, not getting a solution.
    Could you give some hint? 😁
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  • mathbyvemuri

    Member • May 16, 2012

    Dancer_Engineer
    I tried, not getting a solution.
    Could you give some hint? 😁
    HINT:- Today he can give a piece worth a value of 1-day rent. Tomorrow, he can give a piece worth a value of 2-days rent and take back the piece given today, such that it counts for 2-days rent.👍
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  • Anoop Kumar

    Member • May 16, 2012

    ans:5
    1,2,4,8,16 😎
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  • mathbyvemuri

    Member • May 24, 2012

    Yes, as 'ianoop' said, 5-pieces is the answer.
    Beauty of math is illustrated in this problem.
    If Mr X cuts the silver bar of 31 inches in to the pieces with following sizes (in inches), it will result in minimum number of pieces- solution:
    1, 2, 4, 8, 16
    On day-1 Mr X gives 1-inch piece to the landlord
    On day-2 he gives 2-inch piece and takes back the 1-inch piece
    On day-3 he gives 1-inch piece in addition to the already given 2-inch piece, thus making it 3-inches in total
    On day-4 he gives 4-inch piece and takes back the 1-inch and 2-inch pieces
    On day-5 he gives 1-inch piece in addition to the already given 4-inch piece, thus making it 5-inches in total.
    This will continue till day-31.
    Hence a minimum of five pieces is enough
    Math logic:
    Any number up to 2n-1 can be represented by the combinations of the ‘n’ numbers: 2^0,2^1,2^2,...2n-1.
    Answer (A)
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